Question

.. What is the electric potential at the surface of
platinum nucleus ? The radius of nucleus is
6.6 x 10-15 m and atomic number of platinum is 78.
7 τ η​

Answer 1

Answer:

V = 2.56 x {10}^{21} Nm/C

Explanation:

As we know,

$v = k \frac{q}{r} \: \: \: \: \: k = 9 \times {10}^{9}$

Given :

Now the charge on platinum atom is by 78 protons in nucleus So Total charge in nucleus is

$q = 78 \times 1.6 \times {10}^{ - 19} = 1.24 \times {10}^{ - 17}$

q = 1.24 x {10}^{-17} C

Radius, r = 6.6 x {10}^{-15} m

$v = 9 \times {10}^{9} \times \frac{1.24 \times {10}^{ - 17} }{ ({6.6 \times {10}^{ - 15}) }^{2} }$

$v = {2.56 \times10 }^{21}$

V = 2.56 x {10}^{21} Nm/C